题意：由1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1，……合并可得1,22,11,2,1,22,1,22,11,2,11,22,1，再由每个数的位数可得新序列，推出新序列第n项。
分析：新序列与原序列相同，按题意打表即可。
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define lowbit(x) (x & (-x)) const double eps = 1e-8; inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const int MAXN = 10000 + 10; const int MAXT = 10000 + 10; using namespace std; vector v; int main(){ v.push_back(1); v.push_back(2); v.push_back(2); v.push_back(1); v.push_back(1); int cur = 3; while(1){ int cnt = v[cur]; int x; if(v.size() & 1){ x = 2; } else{ x = 1; } for(int i = 0; i < cnt; ++i){ v.push_back(x); } if(v.size() > 10000000) break; } int T; scanf("%d", &T); while(T--){ int n; scanf("%d", &n); printf("%d\n", v[n - 1]); } return 0; }
转载于:https://www.cnblogs.com/tyty-Somnuspoppy/p/7367818.html
